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So there are a few posts about this already, but they skip over the problem I have.

The proof of Cantor's theorem elegantly shows that if we consider $f:A\rightarrow \mathcal{P}(A)$, the set $B=\{x\in A : x\not\in f(x) \}$ by its definition cannot lie in the image of $f$, precluding it from being a bijection - hence power sets have strictly greater cardinality.

But now, I thought one (obvious) way to characterise the "set of all sets" would be $S=\mathcal{P}(S)$, as indeed, it contains all possible sets (perhaps here is the problem, but I don't immediately see how using this as a definition for the set of all sets is bad). Now consider simply taking the identity function $i:S\rightarrow S$. Then $B=\{x\in S : x\not\in x \}$ - the set of all sets which do not contain themselves. But this doesn't exist. It's exactly the set of Russell's paradox. If this isn't a set, and so is not in $S$, then the proof no longer offers an obstruction to $i$ being surjective.

Each other post I've seen on the matter uses Cantor's theorem to show such a set doesn't exist. My query is to find that something else that must be happening, since I think the proof has issues for this particular set. I'm expecting other issues to arise with the definition $S=\mathcal{P}(S)$...

FireGarden
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    My answer here should be useful for this question. Let me know if it answers you question. – Asaf Karagila Sep 21 '15 at 13:57
  • @AsafKaragila It does in a sense. The sentence "The proof of Cantor's theorem that the power set is strictly larger simply does not go through without using "forbidden" formulas in the process." seems to be the problem I've had here, but it's not a comfortable conclusion. It also reads like I would believe Quine's set theory to be 'more correct' for my views.. – FireGarden Sep 21 '15 at 14:11
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    Yes, Cantor's theorem is not true in Quine's NF. – Zhen Lin Sep 21 '15 at 14:18
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    There are people who work with NF and research into its power, but the general opinion (until the proofs are verified and polished) is that NF is quite weak to serve as a strong foundational theory. (I cannot find a comment by Andres Caicedo which compared adding more and more "useful" axioms to NF as working with fragments of ZF), so is it "more correct" than ZFC, who knows. Mathematicians are mostly utilitarians in these things, if it's useful it's good; if it's less useful, it's probably "less good". – Asaf Karagila Sep 21 '15 at 14:40

1 Answers1

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In standard set theory it is a fundamental truth (one of the instances of the axiom schema of separation) that

  1. Whenever $A$ is a set, $\{x\in A:x\notin x\}$ is also a set.

You seem to be thinking that the argument from Russell's paradox shows that this truth cannot hold when $A$ is the set of all sets and must therefore be modified to

  1. Whenever $A$ is a set that is not the set of all sets, $\{x\in A:x\notin x\}$ is also a set.

But this is not actually necessary -- the original (1) works perfectly well, exactly because there is no set of all sets. If we try to let $A$ be a collection of all sets, then the premise "$A$ is a set" is simply false, and therefore it doesn't matter that the conclusion doesn't hold.

We don't usually state the assumption "when $A$ is a set" explicitly because it is implicit in working in set theory that everything we speak about is assumed to be sets. There's a similar hidden assumption in Cantor's theorem:

  1. Theorem (Cantor). Assume that $A$ is a set. Then there is no bijective function $A\to\mathcal P(A)$.

So if you try to apply this to $S$ satisfying $S=\mathcal P(S)$ and get the nonsensical conclusion that the identity function on $S$ cannot exist, then Cantor's theorem still works fine because the thing you're applying it to doesn't actually exist. If you pick the proof of Cantor's theorem apart in this case, you'll find that it uses something much like (1) as the crucial step, indicating that it actually does depend on the assumption "$A$ is a set".

So what you have found is essentially just two ways of phrasing the same argument that there is no set of all sets. They are not in conflict with each other.

hmakholm left over Monica
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